In a triangle ABC, if AD is the angle bisector of ∠BAC, AE ⊥ BC, ∠B = 30° and ∠C = 50°, then ∠DAE is: 

Correct option is C

Given:

△ABC with ∠B = 30$$^\circ$$ and ∠C = 50$$^\circ$$​​

AD is the angle bisector of ∠BAC

AE⊥BC 

Concept Used: ​

Sum of all interior angle of triangle = 180$$^\circ$$​

Solution:​

Finding ∠A in △ABC

∠A = 180$$^\circ$$− (∠B + ∠C)

=180$$^\circ$$​−(30$$^\circ$$ + 50^∘)

= 100$$^\circ$$​​

Since AD is the angle bisector

∠BAD = ∠CAD = $$\frac{∠A}2=\frac{100^∘}{2}$$ = 50$$^\circ$$​​

∠DAE =  90$$^\circ$$- (∠BAD + ∠B) = 90^∘​- 30$$^\circ$$+ 50$$^\circ$$ = 10^o​
​​